NOIP 模拟 $23\; \rm 题$
题解 \(by\;zj\varphi\)
考虑 \(\rm DP\)
设 \(dp_{k}(S)\) 表示前 \(k\) 个人来后 \(S\) 集合中的苹果都存在的概率是否大于 \(0\)
考虑倒着转移
\(\alpha.u_i,v_i\in S\)
\(\beta.u_i\in S,f_k(S\cup\{u_i\})=f_{k+1}(S)\)
\(\gamma.v_i\in S,f_k(S\cup\{v_i\})=f_{k+1}(S)\)
\(\delta.ui,vi∉S,f_k(S)=f_{k+1}(S)\)
最后要修改一些细节,看代码即可
Code
#include<bits/stdc++.h>
#define ri register signed
#define p(i) ++i
using namespace std;
namespace IO{
char buf[1<<21],*p1=buf,*p2=buf;
#define gc() p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?(-1):*p1++;
template<typename T>inline void read(T &x){
ri f=1;x=0;register char ch=gc();
while(ch<'0'||ch>'9'){if (ch=='-') f=0;ch=gc();}
while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch^48);ch=gc();}
x=f?x:-x;
}
}
using IO::read;
namespace nanfeng{
#define FI FILE *IN
#define FO FILE *OUT
template<typename T>inline T cmax(T x,T y) {return x>y?x:y;}
template<typename T>inline T cmin(T x,T y) {return x>y?y:x;}
static const int N=404,M=5e4+7;
int bit[N][N],etn[N],n,m,ans;
struct node{int a,b;}et[M];
inline int main() {
//FI=freopen("nanfeng.in","r",stdin);
//FO=freopen("nanfeng.out","w",stdout);
read(n),read(m);
for (ri i(1);i<=m;p(i)) read(et[i].a),read(et[i].b);
for (ri i(1);i<=n;p(i)) {
bit[i][i]=1;
for (ri j(m);j;--j) {
if (bit[i][et[j].a]&&bit[i][et[j].b]) etn[i]=1;
if (bit[i][et[j].a]||bit[i][et[j].b])
bit[i][et[j].a]=bit[i][et[j].b]=1;
}
}
for (ri i(1);i<=n;p(i)) {
if (etn[i]) continue;
for (ri j(1);j<=n;p(j)) {
if (etn[j]) continue;
ri fg=1;
for (ri k(1);k<=n;p(k)) if (bit[i][k]&&bit[j][k]) {fg=0;break;}
ans+=fg;
}
}
printf("%d\n",ans>>1);
return 0;
}
}
int main() {return nanfeng::main();}